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According to Ampere’s law $$\oint \vec{B}. d\vec{l}=\mu_0I_e$$

Where, $I_e$ is the current enclosed by the amperian loop

So if I consider an amperian loop at the centre of the circular current carrying wire(with the same centre) with a radius lesser than that of the current carrying loop, since there is no current enclosed($I_e$) in the loop there must be no magnetic field at the centre.

But there is a formula for magnetic field due to current carrying arcs which gives us the expression $$\vec{B}=\frac{\mu_0 I \theta}{4\pi r}$$

We can get the magnetic field by putting $\theta=2\pi$

Why is my initial statement wrong?

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    $\begingroup$ Your mistake is assuming that just because an integral vanishes, means that the integrand is zero (most specifically $B=0$), but this isn't true in general. In fact the integral is zero here because everywhere on the loop, B is orthogonal to the loop direction, so the integral is zero by properties of the dot product. $\endgroup$
    – Triatticus
    Sep 2, 2021 at 2:50
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    $\begingroup$ Related question where it mentions the same problem but with Gauss Law being zero and that it does not imply $E=0$. $\endgroup$
    – ludz
    Sep 2, 2021 at 6:08

3 Answers 3

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Ampere's law:

$$\oint_C \vec{B} \cdot d\vec{l}=\mu_0I_e$$

The integral tells us that if the magnetic field is zero at every point along the closed curve C, then there must be no net current enclosed in the loop.

However, the converse is not true -- there are many ways for the magnetic field to be non-zero at most or every point along that curve, and yet still have the total integral evaluate to a total of zero.

In other words, there are many ways to have a strong magnetic field in free space, even when a loop in that region does not include any current.

With an Amperian loop at the centre of a circular current carrying wire with a radius lesser than that of the current carrying loop, yes there is no current enclosed(Ie) in the loop.

However, there is a strong magnetic field pointing through the current carrying loop of wire. When actually doing the integration, the little "dl" vectors are in the plane of the loop of wire, but the magnetic field "B" is at right angles to that plane, so the dot product is zero.

Let's now consider a slightly different Amperian loop, perhaps one still centered on the center of the current-carrying loop of wire and still smaller than that loop, but now tilted at some angle. Since the little "dl" vectors are not exactly at right angles to the "B" field generated by that current, the dot product is not zero, so there is a part of that curve where the integral is positive. However, the integral along the rest of the curve is negative, so the total integral still evaluates to a total of zero.

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In order to go from $\oint \vec B\cdot d\vec \ell= \vert \vec B\vert L=B L$ where $L$ is the length of your contour, you must have that $\vert \vec B\vert$ is the same on all points of the contour, and $\vec B$ is parallel to $d\vec \ell$ everywhere.

Here, this is not the case. If you consider an Amperian loop of infinitesimal radius $\delta r$ in the plane around the center, then $\vec B$ is perpendicular to $d\vec \ell$ everywhere on the contour since, by the RHR, $\vec B$ is completely perpendicular to the plane. Thus $\vec B\cdot d\vec\ell =0$ and $\oint \vec B\cdot d\vec\ell=0$, in accordance with Ampere’s law since the loop encloses no current.

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Actually the statement and converse are both true. If by statement we mean:

$\oint \vec{B} \cdot d\vec{l}=0 \implies $ enclosed current is zero

And converse:

Enclosed current zero $ \implies \oint \vec{B} \cdot d\vec{l}=0 $

This does not imply no field through the surface defined by the loop, just that it integrates to zero around the path of the loop (dot-producted along the way). That’s the issue. You can have lots of field inside the loop, but the vector summation (integral) is zero, which it will be without enclosed current. (I’m just focusing on intuition only.)

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    $\begingroup$ Ampere's law has nothing to do with flux. The integral is a path integral and not a surface integral. There is no surface involved. $\endgroup$
    – nasu
    Sep 2, 2021 at 11:28
  • $\begingroup$ @nasu Youre absolutely right. It is the current density integrated over the 2D surface made by the enclosed loop, ie the enclosed current. $B\cdot dl$ is a path integral. Thanks, and of course edited $\endgroup$
    – Al Brown
    Sep 2, 2021 at 11:41

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